Horvath Practice Problems 75 - This leaves 0.00250.1X mol...

This leaves0.0025−0.1Xmol of niacin and0.1Xmol of C6H4NO−2The concentrations acetic acid and acetate ion are0.0025−0.1X0.025 +Xand0.1X0.025 +Xrespectively.Now we setup an ICE chart[HC6H42]£H+¤£C6H4−2¤I0.0025−0.1X0.025+X00.1X0.025+XC−xxxE0.0025−0.1X0.025+X−xx0.1X0.025+X+xUsing in theKawe can solve forxwhich gives the H+concen-tration,Ka=1.4×10−5=¡0.1X0.025+X+x¢(x)¡0.0025−0.1X0.025+X−x¢.At the equivalence point and equal amount of NaOH has fullyreacted with the niacin.The total volume is now 50 mL and there are 0.0025 mol ofacetate ion, which is0.00250.050=0.05MWe use this in an ICE chart for the conjugate base reaction

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